Brightness of distant galaxies problem

[whiterabbit90]

Homework Statement

An object gives out a spectral line of 919nm compared to that of 589nm in a laboratory. It has a brigtness of 1.2 x 10^-11 W m^-2. The objective is to find out the brightness of the object if it were at 700 Mpc.

Homework Equations

1. Z (redshift)= change in wavelength / original wavelength

2. V= Z X C

3. V= H₀D

4. F= L / 4 pi r²

The Attempt at a Solution

The redshift value I find to be 0.56 using equation 1. Using this I calculate the speed as 168000 km s^-1 using equation 2. I then work out the distance of the object to be 2400Mpc using equation 3 rearranged. now from here do I use equation 4 to calculate the luminosity and then change the distance to 700Mpc or do I divide 2400Mpc by 700Mpc to get how many times closer the object will be square the answer ( 3.43²) and times it by the original brightness 1.2 x 10^-11 W m^-2?

 

[diazona]

Try both ways and see if they give the same answer or not.

 

[whiterabbit90]

I tried both, equation 4 gives me an answer of 4.932 x 10^-9 and using divide and square i get 1.411 x 10^-10 both seem plausible answers.

 

[diazona]

In what units?

If they're the units I suspect you meant, one of those answers is wrong - and not just in the obvious sense that they can't both be right; you miscalculated one of them. Go back and check your math.

 

[whiterabbit90]

both should be in units of W m^-2. In equation 4 I converted Mpc to metres same luminosity is Watts. My equation for 4 looks like this rearranged.

L= F x 4 pi r²

Putting in the values gives 1.2 x 10^-11 W m^-2(F) x 4 pi (7.44x 10²⁵)² m (r²) = 8.35 x 10⁴¹ W (L)

With that I get luminosity to be 8.35 x 10⁴¹ W (L)

Using equation 4 as F= L / 4 pi r2 and using the new value for Luminosity I get.

8.35 x 10⁴¹ W / (4 pi (2.17 x 10²⁵)²)m² = 4.932 x 10^-9 W m^-2

Can anyone tell me where in that I am going wrong as the equations and units all seem to add up if anything I feel this may be the correct answer and the other method using 3.43² may be wrong. However as 3.43² does not have units is shouldn't effect W m^-2.

 

[diazona]

8.35 x 10⁴¹ W / (4 pi (2.17 x 10²⁵)²)m² = 4.932 x 10^-9 W m^-2

This is the problem. The result of that calculation is not 4.932\times 10^{-9}\frac{\mathrm{W}}{\mathrm{m}^2}.

 

[whiterabbit90]

I just calculated it and got 1.411 x 10^-10 I think the problem was I was squaring the entire bottom half of the divide rather than the distance. A stupid mistake but guess we all make them. I assume since both equations give the same answer that is the right answer?

 

[diazona]

Yeah, everybody makes stupid mistakes ;)

Usually when you can do something two different ways and get the same answer, chances are good that you got it right.